 Now the basic definitions of logarithms are all you really need to know to be able to work with any sort of log, however it's convenient to have some shortcut methods, and this leads to what are called the rules of logarithms. And so again, this goes back to our basic definition of what a logarithm is, suppose I have x equals a to the m and y equals a to the power m. Well, the logarithm is an exponent by the definition of what a logarithm is, the log to base a of x is m, here's our base, here's our argument, here's the actual log, base argument, actual log, so log to base a of x is m, log to base a of y is n. By the rules of exponents, if I multiply x times y, they have the same base, so the exponents are going to add together. If I divide them, the exponents are going to subtract, and if I raise one of them to a power, then I'm going to get the exponent itself multiplied by whatever that power is. And because logs are exponents, this gives us a set of rules for working with logarithms. So the log of x times y is the exponent, m plus n, but m is the same thing as log to base a of x, n is the same thing as log to base a of y, so I can replace them, and I have my rule for logarithm of a product. Likewise, if I take a look at a quotient, the log to base a of x over y is the exponent, m minus n, but m is log to base a of x, n is log to base a of y, so that's going to be the difference of the two logs. And finally, the log of a power, x to the k, the log is the exponent, mk, and again m is log to base a of x, this is going to be k times log to base a of x. And so here's our three rules for working with logs, which we often say the log of a product is the sum of the logs, the log of a quotient is the difference of logs, and the log of a power is the power multiplied by the log. For example, let's say I take log to base 7 of x to power 3 times x plus 5 raised to power 5. So a little analysis goes a long way. So the first thing we should do is we should try to determine what type of expression we have, what type of expression is the argument. So let's take a look at that. That's a product of x to the third and x plus 5 to the fifth. So this is a product, and so that says I should apply the product rule for logs. Log of a product is the sum of the individual logs. So my first simplification, log of a product is the log of the first factor plus the log of the second factor. So there's my first step. Now we want to simplify as much as possible because later on this will become really, really helpful. I have my first expression here, x to power 3, that's a power. So I can apply the power rule for log of the power, the exponent becomes a coefficient. So this log to base 7 of x to the third, that exponent 3 is going to become a coefficient, 3 log to base 7 of x. The other term is also a power. So again, I can apply the power rule. That exponent 5 becomes a coefficient, 5 log to base 7 of x plus 5. And what do I have? Well, let's see, can't do much with that. Here I have the log of a sum. And well, x plus 5 is a sum. So I can apply the sum rule for logs, which is log to base a of a sum, is log to base a of the sum. In other words, if I have a sum, I can't simplify it any further. A sum cannot be simplified using logs. And that means I can't do anything else with this. I've already done with that. And this is the, as far as I can go with the simplification. Well, here's another one, log of that horrible thing. And it doesn't actually matter what the base is. The rules of logarithms hold, regardless of what our base actually is. So again, a little analysis goes a long way. What we have is we have a cube root. We're taking the log of something, something, something, cube root of something. And so cube roots, I can express these as powers. This cube root is the same as raising things to power 1 third. So I have a power, and now I can apply the power rule. The log of a power, the exponent comes out front. So this 1 third is going to become a coefficient. 1 third log of 3x plus something over 2x minus 5. And parentheses, like paper, are cheap. So we should go ahead and use them, throw that into parentheses. This is important because it's 1 third is the exponent of the entire expression. So 1 third should be the log of the entire expression. So again, a little analysis goes a long way. This is the log of a quotient. So I can apply the quotient rule for logs. Log of a quotient is the difference of the two logs. So that's going to be log 3x plus 7 minus log 2x minus 5. And again, we're taking the log of a sum and the log of a difference. And they cannot be simplified any further. The log of a sum is the log of the sum. The log of a difference is the log of the difference. And there's nothing else we can do with them. So at this point, we are done with our simplification. Well, let's evaluate a logarithmic expression. So here I want to evaluate log base 2 of 12 minus 2 log 2 base 2 of 3 plus log base 2 of 6. So, well, let's see. I don't know what any of these are immediately. So maybe I can figure something out. Let's analyze the type of expression we have. This is a sum. The last thing we do is we add things together. So this is a sum of logarithmic expressions. And so I have the sum of logs is the log of a product. However, notice that when I add this sum, there's no coefficients here. This is simply the log. Simply the log. I'm not allowed to have coefficients. I can combine them as a product, but I can't have a coefficient there, which means that this 2 is going to be a profit. So I have to do something with that. Well, I can get rid of the coefficient by using the power rule. k log base a of x, that k, that coefficient, can become an exponent. So this expression here, I could rewrite this 2 log 2 base 2 of 3. This is log base 2 of 3 squared. That coefficient 2 has moved inside as an exponent. And now I have a sum and difference of logs. And I can use the product and quotient rules. So this is log of a difference is a quotient 12 divided by 3 squared. And now I have the log of a sum. And I can use the product rule. Sum of logs is the log of a product. And I can do a little bit of arithmetic. And I can do a little bit more arithmetic. And a little bit more. And a little bit more. And I now have log to base 2 of 8. And I can make a final reduction by remembering that 2 to power 3 is equal to 8, or computing that 2 to power 3 is equal to 8. So remember, log is the exponent. The base is the base. The power is the argument. And the actual value of the log is the exponent. And so log to base 2 of 8 is equal to 3.